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设数列{An}(n=1,2,…)是等差数列,且公差为D,...

解: am=Sm-S(m-1)=0-(-9)=9 Sm=(a1+am)m/2=0 a1+am=0 a1=-am=-9 又am=a1+(m-1)d d=(am-a1)/(m-1)=[9-(-9)]/(m-1)=18/(m-1) m>3,18/(m-1)

(1)由an+1=2Sn+3,得an=2Sn-1+3(n≥2)…(2分)相减得:an+1-an=2(Sn-Sn-1),即an+1=3an,∵当n=1时,a2=2a1+3=9,∴a2a1=3,∴数列{an}是等比数列,∴an=3?3n-1=3n…(5分)(2)∵b1+b2+b3=15,b1+b3=2b2,∴b2=5…(6分)由题意,a13+b1,a23+b2...

(1)①当n=4时,a1,a2,a3,a4中不可能删去首项或末项,否则等差数列中连续三项成等比数列,则推出d=0.若删去a2,则a32=a1?a4,即(a1+2d)2=a1?(a1+3d)化简得a1+4d=0,得a1d=?4若删去a3,则a22=a1?a4,即(a1+d)2=a1?(a1+3d)化简得a1-d...

(1)∵数列{an}为公差为d的等差数列,且a1,2a2+2,5a3成等比数列,∴(2a2+2)2=a1?(5a3),即(2a1+2d+2)2=a1?[5(a1+2d)],a1=10,∴4(11+d)2=500+100d,∴d2-3d-4=0,∴d=4或d=-1,∴an=10+(n-1)×(-1)=11-n或an=10+(n-1)×4=4n+6.(2)∵an...

由数列{an}是公差为d,且首项为a0=d的等差数列得:an=a0+(n+1-1)d=(n+1)d;∴Sn+1=a0C0n+a1C1n+…+anCnn,又Sn+1=anCnn+an?1Cn?1n+…+a0C0n=anC0n+an?1C1n+…+a0Cnn,∴2Sn+1=(a0+an)C 0n+(a1+an?1)C1n+…+(an+a0)Cnn=(a0+an)(C0n+C1n+…+Cnn)=...

(1)由题意得a1=1,a2=2,又S3=a4,a3+a5=2+a4,∴a1+a2+a3=a4a3+a5=2+a4,∴1+2+1+d=2q1+d+1+2d=2+2q即4+d=2q2+3d=2+2q解得d=2,q=3;(2)当n为奇数时,sn=(a1+a3+…+an)+(a2+a4+…+an-1)=n+12(a1+an)2+a2(1?qn?12)1?q=n+14[1+1+(

∵a1=23,公差d=-2∴an=23-2(n-1)=-2n+25≥0,∴n≤252即前12项是正数,∴其前n项和Sn达到最大值时n为12故选C.

解:(Ⅰ)由题意知:d>0, , , ,化简,得 , ,当n≥2时, ,适合n=1情形,故所求 。(Ⅱ) , 恒成立,又m+n=3k且m≠n,∴ ,故 ,即c的最大值是 。

(1)①由题意,得2a1+4d=178a1+28d=56,解得d=-1…(4分)②由①知a1=212,所以an=232?n,则bn=3n?an=3n?(232?n),…(6分)因为bn+1-bn=2×3n×(10-n)…(8分)所以b11=b10,且当n≤10时,数列{bn}单调递增,当n≥11时,数列{bn}单调递减,故当n...

(1)∵{an}是公差为d的等差数列,∴am=a1+(m-1)d,an=a1+(n-1)d,am+an=2a1+(m+n-2)d,又m+n=2p,∴am+an=2a1+2(p-1)d,∵a1+(p-1)d=ap,∴am+an=2ap. …(3分)∵{bn}是公比为q的等比数列,∴bm=b1qm-1,bn=b1qn-1,bmbn=b12qm+n-2,∵m+n=2...

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